Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question that asks to find all local extrema for the following function $$f(x)=x^3-\sqrt{4x}$$

I started by finding computing the first derivative: $$f'(x)=\frac{3x^{5/2}-1}{\sqrt{x}}$$ $f'(x) = 0$ if $x=\left(\frac{1}{3}\right)^{2/5}$. Also, $f'$ is not defined when $x=0$. In finding the local extrema do I include $x=0$?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Yes, you need to add $0$.

Study the variations of $f$ to convince yourself. It is nonincreasing on $[0,x]$ and nondecreasing on $[x,+\infty)$.

So $f(x)$ is actually a global minimum.

And since $\lim_{+\infty}f(x)=+\infty$, $f(0)$ is a local (not global) maximum.

Be careful, not every critical point is a local extremum. You need the first derivative test or the second derivative test, for instance, to conclude. That's why, anyhow, the best way is to fully study the variations of the function on its domain by determining the sign of its derivative, when it is available like here.

enter image description here

share|improve this answer
    
Thanks for your response. So in terms of local extrema, $f(x)$ has a local minimum at $x=\left(\frac{1}{3}\right)^{2/5}$? –  Paul Mar 6 '13 at 0:40
    
@Paul Yes. And it is even global. See also my edit about critical points. –  1015 Mar 6 '13 at 0:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.